Question

The position of a particle is expressed as r=(4t^2i + 2tj) m, where t is time in second. Find the velocity of the particle at t=3s. ​

Answer 1

Answer:

$v = 8t \: i + 2 \: j \\ \\ at \: \: t = 3 \\ \\ v = 24i + 2j = \sqrt{ {24}^{2} + {2}^{2} } = \sqrt{580}$

Answer 2

Answer:

from the position of a particle the velocity of the particle at t=3s

is OPTION A).24.083 m/s

Explanation:

given that position vector is

$r = 4 {t}^{2} i + 2t \: j$

then to find velocity

$v = \frac{d}{dt} (r) = \frac{d}{dt} (4 {t}^{2}i + 2t \: j ) \\ implies \\ v = 8t \: i \: + 2 \: j \\ at \:t = 3 \: sec \\ the \: velocity \: is \: v = 24 \: i + 2 \: j \\ then \: the \: magnitude \: of \: \\ the \: velocity \: vector \: is \: \\ v = \sqrt{ {24}^{2} + {2}^{2} } = \sqrt{580} = 24.083.$

the velocity of particle with position vector r = 4 {t}^{2} i + 2t \: j \\ at t=3s is A).24.083 m/s.

the velocity of particle with position vector at t=3s is A).24.083 m/s.#SPJ3.